Compound events and simulations: a week of ready-to-teach maths

Five lessons for Year 8 Probability. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus a class data set, graphs to plot on, a random digit table, a printed spreadsheet run, door cards for a game, a mini-check and every answer with its working.

Start here: five minutes to Monday

1. Skim the week at a glance on the next page, and book computers or tablets with a spreadsheet for Day 4.
2. Print the five days. Each day is two A4 sheets: a plan and a worksheet. The data sets and graphs are on the worksheets.
3. Print the two cut-out sheets, one copy per pair: the random digit table (Day 3) and the door cards (Day 5).
4. Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.
5. Teach straight from the plan. Timings, talk prompts, misconceptions and worked answers are all on the one page.

The idea in plain words

A compound event is made of more than one result, such as two heads when two coins are tossed, or at least one six when two dice are rolled. Its theoretical probability comes from counting equally likely outcomes. Its experimental probability comes from doing it many times: the number of times it happens divided by the number of trials. One short experiment wanders, but repeating and pooling steadies the estimate. When doing it for real is slow, or counting is too hard, a simulation stands in: random digits or a spreadsheet play the part of the coins, dice or shots, thousands of times if we like. Describing a result means giving the estimate, the number of trials it came from, and how it compares with the theory.

One day, one lesson

The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.

Using real results

Days 1 and 2 use the class’s own coins and dice. Each worksheet also prints a set of results made the same way, so a lesson still works when time runs short, and the answers are worked from those printed results.

The week at a glance

One lesson a day for a week. Each day stands on the day before, so run them in order.

How the week builds

Day 1 sets the theory for a compound event and runs a first experiment; Day 2 repeats and pools; Days 3 and 4 simulate, first with random digits and then with a spreadsheet; and Day 5 describes results. It builds on Year 7 experiments and simulations with single events and on combined events earlier in Year 8 (tables, trees and Venn diagrams), and prepares compound events with and without replacement, and simulations that compare simple and compound events, in Year 9.

Materials for the week (one trip)

• From the classroom: pencils, rulers, calculators, scissors, this pack printed.
• Two different coins per pair, such as a 10c and a 20c (Day 1), and two dice per pair (Day 2).
• Cut-out sheet 1, one per pair: the random digit table (Day 3).
• A computer or tablet with a spreadsheet, one per pair (Day 4).
• Cut-out sheet 2, one per pair: three door cards (Day 5).

Words for the word wall

compound event, outcome, equally likely, theoretical probability, trial, experimental probability, relative frequency, tally, pool, estimate, simulation, random digit, spreadsheet, formula, recalculate, variation, long run

Dear families

This week in maths, Year 8 is finding the chances of events made of more than one result, such as two heads from two coins. They work out the chance by counting, then test it by experiment and by simulation on a spreadsheet, and learn to describe how close the two are.

Try this at home

• Toss two coins 20 times together. How often do they match? (Counting says about half the time, 10 of 20.)
• Roll two dice 36 times. How often is there at least one six? (Counting says 11 of 36.)
• Ask: is 7 heads in 10 tosses strange for a fair coin? (Not very: short runs often stray from half.)
• If you use a spreadsheet, type =RANDBETWEEN(1,6) in a cell and fill it down: a whole column of dice rolls.

My experiments this week

Fill one row a day. Name the compound event, the number of trials, and what you found.

Printed from the free seegongsik Compound events and simulations teaching pack · seegongsik.com/au/y8/probability/AC9M8P03/pack

Compound events

A compound event is made of more than one result. Students list the four outcomes for two coins, find theoretical probabilities, predict how often each event should happen in 40 tosses, then toss two coins 40 times and compare.

We are learning to

• recognise a compound event and list the outcomes that make it happen,
• find its theoretical probability by counting equally likely outcomes,
• predict how many times it should happen, and compare with what does happen.

Success criteria

• I can explain why HT and TH are different outcomes.
• I can predict about 10 lots of two heads in 40 tosses, and say why my result may differ.

You need

Two different coins per pair, such as a 10c and a 20c; the worksheet each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the 40 tosses.

Watch for these ideas

• Listing HT and TH as one outcome, which makes two heads look like 1/3.
• Expecting the result to match the prediction exactly.
• Reading “at least one head” as “exactly one head”.

Answers

• Part A: HH, HT, TH, TT. two heads HH: 1/4, about 10 in 40; a match HH, TT: 2/4 = 1/2, about 20 in 40; at least one head HH, HT, TH: 3/4, about 30 in 40.
• Part B: results vary from pair to pair; the counts add to 40.
• Part C, the printed run: two heads 10/40 = 0.25 (theory 0.25); a match 24/40 = 0.60 (theory 0.50); at least one head 26/40 = 0.65 (theory 0.75). Close, but not equal: a short run wanders.
• Part D: 8 outcomes, HHH, HHT, HTH, HTT, THH, THT, TTH, TTT; all match 2/8 = 1/4, so about 10 in 40.

Compound events

A. Theory first

List the four outcomes of tossing a 10c and a 20c (10c first): ____________________________. Then fill the table.

B. Toss and tally

Toss both coins together 40 times. Make a tally mark for each toss.

C. Compare

Use your tallies. If you have none, use this printed run of 40 tosses: HH 10, HT 7, TH 9, TT 14.

Which event came closest to the theory? Why might none match it exactly?

D. Stretch: three coins

List the outcomes when three coins are tossed. What is the chance all three match? About how many times in 40 tosses?

Repeating and pooling

One pair’s result wanders; pooling many pairs steadies it. Students roll two dice 30 times and count the rolls with at least one six, compare twelve pairs on a dot plot, and graph the pooled estimate as each pair is added.

We are learning to

• find an experimental probability as a relative frequency,
• describe how results vary from one short experiment to the next,
• pool results and explain why the pooled estimate is more reliable.

Success criteria

• I can find 11/36 for at least one six by counting, and compare it with my result.
• I can explain why 360 rolls give a better estimate than 30.

You need

Two dice per pair; the worksheet and a calculator each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the class data.

Watch for these ideas

• Counting the roll 6 and 6 twice, which gives 12/36.
• Taking one pair’s result as the true probability.
• Expecting the pooled line to land exactly on the theory.

Answers

• Part A: 6 rolls with a six first, 6 with a six second, and 6 and 6 is in both, so 6 + 6 − 1 = 11: 11/36, about 0.31; expected about 9 in 30 (30 × 11/36 = 9.2).
• Part B: 0.40, 0.33, 0.23, 0.27, 0.23, 0.40, 0.30, 0.30, 0.20, 0.40, 0.27, 0.20; from 0.20 to 0.40, so pairs differ by up to 6 rolls in 30.
• Part C: 0.40, 0.37, 0.32, 0.31, 0.29, 0.31, 0.31, 0.31, 0.30, 0.31, 0.30, 0.29. It starts at 0.40, swings less with each pair, and ends at 0.29, near 0.31.
• Part D: the pooled 360 rolls; one pair of 30 rolls can land far from the theory, as the dot plot shows.

Repeating and pooling

A. At least one six

B. Twelve pairs

A class of twelve pairs each rolled two dice 30 times. Find each relative frequency, then make a dot plot of the counts.

C. Pool the class

Add one pair at a time. Pooled estimate = all the sixes so far ÷ all the rolls so far. Plot each one.

D. Stretch: which to trust?

Which is the better estimate of the probability: one pair’s result or the pooled class result? Use the dot plot and the graph.

Simulating with random digits

When doing the real thing is slow, random digits can stand in for it. Students simulate a goal shooter’s two shots with pairs of digits, check the result against the 100 equally likely pairs, then simulate collecting a set of four toys, a problem that is hard to count.

We are learning to

• choose digits so that each result has the right probability,
• run and record trials of a compound event from a random digit table,
• compare a simulated estimate with the theory, or use it when the theory is too hard.

Success criteria

• I can explain why the digits 0 to 6 give a 70% chance.
• I can run 20 trials of two shots from my row and find the relative frequency.

You need

Cut-out sheet 1 (the random digit table), one per pair, each pair starting on a different row; the worksheet each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with the toys.

Watch for these ideas

• Using the digits 0 to 7 for 70%: that is eight digits, 80%.
• Reading one digit per trial instead of two.
• Stopping a toy trial at the fourth box instead of when all four toys have come.

Answers

• Part A: rows vary. Row 4 starts 13 71 61 14 67: 3 of these 5 are two goals, and the whole row gives 14 of 20. All 20 rows together: 202 of 400 = 0.51.
• Part B: both digits 0 to 6 in 7 × 7 = 49 of the 100 pairs, so 49/100 = 0.49.
• Part C: from row 4 the first trials take 10, 11, 7, 8 boxes. In the long run it takes about 8.3 boxes, and all four come within 6 boxes about 38% of the time: counting that exactly is beyond Year 8.
• Part D: an even digit for right, five digits a quiz; 4 or 5 right in 6 of the 32 answer patterns, 0.19. The whole table: 25 of 160 = 0.16.

Simulating with random digits

A. Two shots at goal

A goal shooter scores 70% of shots. Digits 0 to 6 are goals; 7, 8 and 9 are misses. One trial is a pair of digits: two shots. Read your row of the table in pairs, and tick each pair that is two goals.

B. Check with the theory

C. Collecting four toys

Each cereal box holds one of 4 toys, all equally likely. How many boxes until you have all four? Digits 1 to 4 are the toys; skip 0 and 5 to 9. Carry on along the table from where you stopped, and do four trials.

Class estimate: P(all four within 6 boxes) = ______   Average number of boxes = ______

D. Stretch: guessing a quiz

Kim guesses every answer on a quiz of 5 true-or-false questions. Design a simulation with random digits for the chance of 4 or more right. Run eight trials. How could you work out the theory?

Simulating with a spreadsheet

A spreadsheet runs thousands of trials of a compound event in moments. Students build a two-dice spreadsheet for at least one six, finish a printed run by hand to see what the formulas do, and compare repeated runs of 20, 200 and 2000 rows.

We are learning to

• write spreadsheet formulas that roll two dice and mark a compound event,
• find an experimental probability from a column of 1s and 0s,
• describe how the spread of results shrinks as the runs get longer.

Success criteria

• I can explain what =IF(OR(A2=6,B2=6),1,0) does.
• I can say why five runs of 2000 rows agree more closely than five runs of 20.

You need

A computer or tablet with a spreadsheet per pair; the worksheet each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with the runs.

Watch for these ideas

• Treating a recalculated run as a mistake, instead of as fresh trials.
• Writing AND where OR is meant: AND marks only the roll 6 and 6.
• Reading one long run as exact, rather than very close.

Answers

• Part A: estimates vary, but 2000 rows should land near 11/36, about 0.31.
• Part B: C is 1 in rows 8, 10, 18, 20 and 0 elsewhere; 4 of 20 = 0.20.
• Part C: ranges 20 rows 0.30, 200 rows 0.09, 2000 rows 0.02: the longer the runs, the closer they agree, all near 0.31.
• Part D: C2 =A2+B2, D2 =IF(C2=7,1,0), then =AVERAGE(D2:D2001); theory 6/36 = 1/6, about 0.17.

Simulating with a spreadsheet

A. Build it

1. In A2 type =RANDBETWEEN(1,6), and the same in B2: one roll of two dice.
2. In C2 type =IF(OR(A2=6,B2=6),1,0): 1 when there is at least one six, 0 when not.
3. Copy A2:C2 and paste it into A3:C2001. That is 2000 rolls.
4. In E1 type =AVERAGE(C2:C21), in E2 =AVERAGE(C2:C201) and in E3 =AVERAGE(C2:C2001): estimates from 20, 200 and 2000 rolls.
5. To make a new run, type anything in an empty cell and press Enter: the spreadsheet recalculates every roll.

My first run: E1 = ______   E2 = ______   E3 = ______

B. A printed run

These are the first 20 rows of one run. Fill column C as the formula would, then find the estimate.

=AVERAGE(C2:C21) = ______

C. Short runs and long runs

Five runs at each length, from one spreadsheet. Find the range of each row.

Make five new runs of your own. What happens to the spread as the runs get longer? Where do the long runs sit?

D. Stretch: a total of 7

Change the spreadsheet to estimate the chance that two dice total 7. Which formulas change? Compare with the theory.

Describing results

A result is only as good as its description. Students play the three-door game, pool the class, explain the result by listing equally likely cases and describe it, then play the self-check quiz as a class game.

We are learning to

• run a game as a repeated experiment and pool the class results,
• explain a result by listing equally likely cases,
• describe a result with its estimate, its number of trials and the theory.

Success criteria

• I can explain why switching wins two times in three.
• I can write a result that says how many trials it came from.

You need

Cut-out sheet 2 (three door cards), one set per pair; the worksheet each; the self-check quiz at the bottom of the unit page for the game.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with Part C and the quiz.

Watch for these ideas

• Thinking two doors left means a 50-50 chance: the host’s choice depends on where the prize is.
• Judging from one pair’s 10 rounds.
• Describing a result without saying how many trials it came from.

Answers

• Part A: results vary from pair to pair.
• Part B: sticking won 41 of 120 = 0.34; switching won 77 of 120 = 0.64. Pairs 3 and 4 did as well or better by sticking: 10 rounds is a short run.
• Part C: prize behind 1, host opens 2 or 3: sticking wins; prize behind 2, host opens 3: switching wins; prize behind 3, host opens 2: switching wins. Sticking wins in 1/3 of the cases and switching in 2/3.
• Part D: for example, “In 120 rounds, switching won 77 times, 0.64, close to the theory of 2/3.” More rounds would make the estimate steadier.

Describing results

A. Play the game

The host shuffles the three door cards, peeks, and lays them face down as doors 1, 2 and 3. The player picks a door. The host turns over a different door that is empty. The player sticks or switches, then turns over the chosen door. Play 10 rounds sticking and 10 switching, taking turns as host.

B. Pool the class

A class of twelve pairs played 10 rounds of each. Find the totals and the experimental probabilities.

P(win by sticking) = ______   P(win by switching) = ______

C. Explain by listing

You pick door 1. The prize is equally likely to be behind each door. Fill in the table.

So P(win by switching) = ______ . Does the class result agree?

D. Stretch: say it so it can be trusted

Write two sentences a newspaper could print about the class result. Give the estimate, the number of rounds, and the theory. How many rounds would make you more sure, and why?

The random digit table

For Day 3, one per pair

Every digit from 0 to 9 is equally likely in each place, and each digit has nothing to do with the one before it. Start on the row your teacher gives you, and read along the row; at the end of a row, carry on with the next.

Teacher note: give each pair a different starting row, so the class pools different trials. The table was made by a computer’s random number generator.

Three door cards

For Day 5, one set per pair

Cut out the three cards and fold each one in half along the dashed line, picture inside. To play, the host lays the folded cards in a row as doors 1, 2 and 3. To open a door, unfold its card.

Keep the cards folded between rounds, and shuffle them out of the player’s sight.

What we know: Compound events and simulations

Work on your own. Show your working, and give probabilities as fractions or decimals.

1. Two coins are tossed. List the four outcomes. What is the theoretical probability of exactly one head? ______________________
2. A coin is tossed and a die is rolled. What is the theoretical probability of a tail and an odd number? About how many times would you expect it in 60 trials? ______________________
3. Two dice were rolled 50 times and a double came up 11 times. What is the experimental probability of a double? What is the theoretical probability? ______________________
4. Ana estimated P(at least one head with two coins) from 20 trials and got 0.65. Ben used 400 trials and got 0.74. Which estimate is more reliable, and why? ______________________
5. A shooter scores 60% of shots. With the digits 0 to 5 as goals, use 38 71 05 92 64 41 to simulate 6 pairs of shots. What fraction of the pairs are two goals? ______________________
6. A spreadsheet rolls two dice in A2 and B2. What does =IF(AND(A2=6,B2=6),1,0) record? After 3,000 rows, what should the average of that column be close to? ______________________
7. Three spreadsheet runs of 1000 rows estimated P(a total of 7) as 0.171, 0.158 and 0.165. Describe these results. ______________________
8. Lee says, “We got 3 heads in 10 tosses, so this coin must be unfair.” Write a reply. ______________________

Answers and marking guide

Answers with working

1. HH, HT, TH, TT; exactly one head is HT or TH, 2/4 = 1/2.
2. 12 outcomes; a tail with 1, 3 or 5 is 3, so 3/12 = 1/4; about 15 in 60.
3. 11/50 = 0.22; theory 6/36 = 1/6, about 0.17.
4. Ben’s: 400 trials vary much less than 20, and 0.74 is close to the theory of 3/4.
5. 38 not, 71 not, 05 two goals, 92 not, 64 not, 41 two goals: 2/6 = 1/3 (the theory is 36/100 = 0.36).
6. A 1 only when both dice show 6, a double six; close to 1/36, about 0.03.
7. All three are close to the theory of 6/36 = 1/6, about 0.167; they differ a little because each run is new random trials.
8. Ten tosses is a short run: a fair coin gives 3 or fewer heads about 17% of the time. Toss it many more times before deciding.

A quick three-level guide

Eight questions, four ideas. A student at standard answers most questions with working and says how many trials each estimate came from.

Weekly class record

Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.

The five columns are the five days: compound events, repeating and pooling, random digits, spreadsheets, and describing results.